3. The attempt at a solution
I figured for a single particle the kinetic energy is:
\begin{equation}
T = \sum_{i=1}^{3} \frac{1}{2} m (\delta_{t} u_{i})^{2}.
\end{equation}
The potential energy should be some force constant times the displacement squared:
\begin{equation}
U = \sum_{i=1}^{3} k (a \delta_{i}u_{i})^{2}.
\end{equation}
So for N\end particles the Lagrangian would become:
\begin{equation}
L_{discrete} = \sum_{n=1}^{N} \sum_{i=1}^{3} \left[ \frac{1}{2} m (\delta_{t} u_{i})^{2} - \frac{k}{3} (a \delta_{i}u_{i})^{2}\right].
\end{equation}
Where the factor 1/3 comes from the fact that I am counting each bond three times.
Now this is where it gets a bit fuzzy for me; what exactly does it mean to take the continuum limit? I believe it has something to do with letting m and a approach zero in such a way that if N approaches infinity the mass stays constant, but I'm not exactly sure how to do that/what that means.

I believe that I require the fact that this is a cubic lattice to be able to write the potential energy down the way I did. But how would change the potential energy if, say for instance, the particles were on a face centered cubic lattice? Or any other lattice for that matter.